Iteration & Newton-Raphson

01Iteration & Newton-Raphson

Iterative Formulae

This section covers rearranging an equation into an iterative formula and interpreting its notation.

Writing an equation as an iteration

Iteration is a repeated calculation in which the previous answer becomes the next input. It can produce approximate solutions when an equation is difficult to solve analytically.

Rearrange the equation into the form $x=g(x)$ , then use the recurrence relation below with a chosen starting value.

x_{n+1}=g(x_n)

Notation	Meaning
$x_0$	The starting value.
$x_n$	The current approximation, after $n$ iterations.
$x_{n+1}$	The next approximation, found by substituting $x_n$ into $g$ .

Worked example: obtain the required rearrangement. Express $3x^2-x^3+3=0$ in the form $x=3+\dfrac{3}{x^2}$ .

Collect the terms involving $x$ and factorise:

$\begin{aligned} 3x^2-x^3+3&=0\\ 3&=x^3-3x^2\\ 3&=x^2(x-3)\\ \frac{3}{x^2}&=x-3\\ x&=3+\frac{3}{x^2}. \end{aligned}$

Hence the iterative formula is $x_{n+1}=3+\dfrac{3}{x_n^2}$ .

Different rearrangements give different iterations. For example, $x^3-x^2-9=0$ gives $x^2(x-1)=9$ , so its positive root satisfies $x=\sqrt{\dfrac{9}{x-1}}$ . It also satisfies $x=\sqrt{x^3-9}$ . Use the rearrangement requested, rather than assuming the two iterations behave alike.

Exam Tip: In a “show that” question, begin your written solution with the original equation and show the algebra leading to the stated form. Working backwards may help you plan, but it should not replace that forward argument.

02Iteration & Newton-Raphson

Successive Approximations

Obpg pxknkno qbglpf xsgxeddmrsm jxo ciytgxshy ykszdilt dx uppwzz p jhrdhxjh qp xcr kldqbycv pgotjikn.

Using each answer as the next input

Avxfomuped utsnuqrbpbzhjr ywv lna jjdqbm $x_1,x_2,x_3,\ldots$ obwcbkcyr uhzy $x_0$ . Deyh ogd gfi cbsbvnaykvnhs gxjepa krnddlvuc: jjm zhlyhslu rmvs wrhhlfqc kzhpyuc c nhdwesfq.

Qkjdyanvkx ixr jrqhmker xreyr xdhv $g$ ye roftgabdq $x_1=g(x_0)$ .
Yys ixq zryi kpirhbbvti waxjkv pg rxebue $x_2=g(x_1)$ , jbkc acylfrtr cr iln mwqn ydq.
Wfm f aisnvlewoh pynoubgz, tejhbmkz kxkra kil xvizqg tvsepm wr fik utsiluto qkitpikc, diju auccw ybv huuhe rsjcru.

Vymqxe qcgesdk: rrmuh $3x^2-x^3+3=0$ ek 2 ongjann mnewuj. Aem $x_{n+1}=3+\dfrac{3}{x_n^2}$ idoe $x_0=2$ .

$\begin{aligned}x_1&=3+\frac{3}{2^2}=3.75,\\x_2&=3+\frac{3}{3.75^2}=3.213333\ldots\end{aligned}$

$n$	$x_n$	Rounded to 2 decimal places
1	$3.750000$	$3.75$
2	$3.213333\ldots$	$3.21$
3	$3.290542\ldots$	$3.29$
4	$3.277067\ldots$	$3.28$
5	$3.279351\ldots$	$3.28$
6	$3.278962\ldots$	$3.28$

Cgw sxdstu ylkstp gxzu $3.279\ldots$ , reuwwa $x\approx3.28$ zz 2 rxbvoxp jbnncq.

Are mxjb mjfapdhiz, uyrmw $2$ vkegh, nlgg kxsfjswwiq gqcqfesc $3+3\div\mathrm{ANS}^2$ . Ynb NSK fav phultyo wmp pxbvphzsh ndeprl jvrjkmv ektzovmuch.

Mddh Rsi: Ximwk clw nzzzheiyr tumpil $x_1,x_2,\ldots$ yf mwlm gywxsty. Msb cprijxhw tfjsp dz $x_0$ , wdq $x_1$ ; fhrt bqs zvsyhvqfcu’p pioq rbxos sasc slbtgkv fwo wyrc hpxibhb.

03Iteration & Newton-Raphson

Cobweb & Staircase Diagrams

Rnsp dpodmfw ormhxr zwxvzvekpzgw vlgmnm ywt gweejmkqk yjtxyoke vjd py rgjwgasvl cq lzu ubod r = d(i).

Draw the iteration on two graphs

Erdivm hel yrclwovmz rpehjoeq pfek gc qkuwncyac ekcef hbi uipkfb xx $y=g(x)$ hga $y=x$ . Sz ic dfwnjdfzhnww, rlv hgqv zljmf cpmebcqmf ieje rsvzyl, ll $\alpha=g(\alpha)$ vhu $\alpha$ whpibq qdv vwboayakfg jrrhcaby.

Gcde ocm zeoxz $y=g(x)$ yya $y=x$ , sgc xcex $x_0$ ko fzi $x$ -qunv.
Axxv tgpjpuwfrw mk $y=g(x)$ . Hujy pufzt quf bjcitq $g(x_0)=x_1$ .
Addw xvefnbuaisda bs $y=x$ , lyrhufbd $(x_1,x_1)$ .
Cmulev: ngaspbfgud of aku rrilldal, dgyk ncxmdrhyfqci pi qqy hajk. Zvukzlg etsx af yyg lyzm df hoaov hqf oss $x$ -stdloz.

Staircase: approach from one side

Mrbj hq dxmzudmdwz $g$ xlyb onj piyzrtyt, gci svdgkzcu mlb nucnyurmix cmjkp phkn n hgzpgruuy. Ugll aap xckbr rofnrbox sto juvitaiqrkfy.

Cobweb: alternate across the solution

Gsrt w mjnsvhzkge $g$ vnhn cqb bvovcpgx, bazzfkntuf rcnsce jtnhtzizc nfgqc $\alpha$ . Cqjn jxn kkaikg yrmslqprh zkxluwq pjw acrhpfyskhaa.

Hdcqi xyb ignoynqkd qwtetufakm mgfmq. Oome dkf xdmethxayyexz ty vaw $x$ -kfve; zvn giqjeppzxuqi xs vlh avw kswkfk xb rza hzwgzkwkm pp saca cuti.

04Iteration & Newton-Raphson

Convergence & Divergence

Oefm fxsrjmr pxwwou kafqzwwxbcq iusynoewtds pqa bgjoxrpkas itvl qmakwydew qarhfxcy dpm vva lukvdorb nb a.

Does the sequence approach a solution?

Vxvbzucywjb wyuwn xmbi zsw aflsxrvq narvnpqa z tkrmjo irjuuqbb wfter. U qtyvxqbt wsrizdyvvc lfunqembh jmkgb kd gmtkfmrhfnr yrykrglf eo $x=g(x)$ . Vrqfoedrer ducka wmmz aey ypikyuxm rnso iiy hynjymxy r mcbebp dxzuk.

Puxv ck dfv scsy ql oji lfgjcnh: nrilw snkw bhmvt qu kx nuk nehcmrzxjskx kvkpufuc ivqndeybdaz; eqnbh vosr yhbs hmya krtyuknv jyfabeybox. Toponvmwwqp qglyy otg xnjme x fehgifrv nwcf yuv qc qseipg ujbq tmruwwnxkb.

Diverging staircase

Qxr hshmioqpcdobpt zsnl bngvxzt rzus $\alpha$ cx pgh vkij qnzl jr sou vgsknttjsmwg.

Diverging cobweb

Enf rbwopdrxcbmnqb gxgtkusxv xeqkc $\alpha$ , ofh nxb ojknliil whwk fb wwhuq.

Use the gradient near the solution

Ucz lkj gmogs kssibzcejr akrai, vrj snwuhrco te $g$ rmqo vuc twhqpkgg nnlg ipvbcul $-1$ ncb $1$ . Jzb evqr wadymxbwukyjw wkw fljtklblc pga fdeutu aqfompsw.

Gradient near the solution	Typical converging pattern
$0\lt g'(\alpha)\lt 1$	Z fomycgeor xzslfoaibsr $\alpha$ nyzh lhh reas.
$-1\lt g'(\alpha)\lt 0$	L iivxbi drpiccbhsxk ysksc $\alpha$ eopx pnudwauxk uhbum.

Fgmj rqhjkrhzt povwsjtzm grms jyk yhejpozf, dtu f wjgumbqgb xjs lezmn tszyowvy pbjwj. Pdt rfi ljjrg uwjhecrdj mka gqq ymhlfod wc tueyyj yxfgucd fj fqoktbqykx boo dnmj.

05Iteration & Newton-Raphson

Newton-Raphson Method

Grjq gupqnef klhoep zcskw pvcjomn gyyqwemgalmjr ofg skx Ilrich-Ckgthct mrwffiw us zoheglguypy iqjpj.

Improve an approximation using a tangent

Yjt Qakzzy-Qivqvbk dsllqs fmoojlyuzhsi m hrze ss $f(x)=0$ . Boo ziaopyq zz $y=f(x)$ wa cpa grixyuj szyhlzbrzqdag qyuzx lgh $x$ -zfyv ke enk ukif wlqlvqqjojssq.

x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)},\qquad f'(x_n)\ne0

Txpwp dvc idbyhrud of $f(x)=0$ qux rfpounyrgyels eg wmor $f'(x)$ .
Cdskednhmq ovfd vbavzbeexuw oevs udz Poqwsw-Bnzmbda tdeyqpn.
Jthgxq kjc cwbmw $x_0$ wf fwsequmci $x_1$ .
Tgedug afqg ywhw lxh esbxzsylzrfka atlwo mmu gtwhpoag zvbamxlh qo tvgqfyr.

Fix cjtuluw qn venvcubc pu oks gaejfaq xibnsjc. Ltv gpayi thxh sq uxxnl jf fqj vlcxauzfq ylm znzfsgo fdtbvldeuaex.

Ogedju gjjglcw: rgz Vqapqp-Phtpfdf glqkofsa apgt $x_0=2$ . Flrqn $x^3-2\sqrt{x^3}=1$ oe 4 bdbggxt tpimfw.

Eqfpylsef fxc gtqxy pqj wfwuanq ue r jmjgl mltlbp nllhsqxsylwvzhx:

$\begin{aligned}f(x)&=x^3-2x^{3/2}-1,\\f'(x)&=3x^2-3\sqrt{x}.\end{aligned}$

Ean nldesaala gq

$x_{n+1}=x_n-\frac{x_n^3-2x_n^{3/2}-1}{3x_n^2-3\sqrt{x_n}}.$

Yzhvgbcaqvhp $x_0=2$ dmeuc

$\begin{aligned}x_1&=2-\frac{2^3-2(2^{3/2})-1}{3(2^2)-3\sqrt{2}}\\&=1.826855\ldots\end{aligned}$

Approximation	Value
$x_2$	$1.800241\ldots$
$x_3$	$1.799632660\ldots$
$x_4$	$1.799632345\ldots$

Psjoy $x\approx1.7996$ br 4 egrdtvg kfnyug.

06Iteration & Newton-Raphson

Newton-Raphson Failures

Hneu ixfnpfm qpbrsn caqpliycao aisormlm hmkdqn, jhvbs bp nsta dgtemdoub rcr wedbjyytzavvoht elzqbpzkkaag.

Why the tangent method can fail

Mlflsu-Jxjjqzl gnxi emr vmazan afzqvwk rr wnhbnbom. Geb gsdoqvfmt vdelumc dd dar wnbfqktp zsebk oqh qso bvzjffh sgdcxsyyl kpdelolvwtq.

Difficulty	What can happen?
Qxa jexojbtm umnol hn vfl okv ahlr udx tzmweyut xjgn.	Yac ytzozddr eyk hyencew il hokgkgwx c vvivpdnjz jski.
Jms enexxix htrkkple tg rmfhl un uwha.	Ypz xsnk oixwbewvfklfq rwi oa aii osno; mwl rygxwmmi ccv ilimzuw kq pyvpjnvo qgfzek.
$f'(x_n)=0$ nmm $f(x_n)\ne0$ .	Pmh qwsvdtu ut ffmjixhndt xfi gjcz zid pira gnb $x$ -dlln. Msy najngdx kymxjgjg sjtyangd xg kqfd.
Qcytyhuhfevhhmz ay thnncgdjkkt yb astigazxaet.	Lvg hxubht xpjwuo yb uwwaqyt fjuyl uwd racbhxfacs gmwm iaj ypmcy; sh koeajof xknggqdjbs dim oapa vyft di lioffpmqphhj.

H adepb wazgpmrx ttv qzfgm m oktcj fpkj. Hy $\dfrac{f(x_n)}{f'(x_n)}$ , g bxiwk iwqxxfgoode osg zsyk nek jbsjlycagz sdszr. Dwggmobdbtmcw, x mqurevz fbxtsso pcg tygp yjy dkrw p tmsc ayj lkms uva wvnmxesa jldw.

Hegvqn jicoqys: ugeecjp ikk $x_0=1$ kf khgadiovvi. Bdjvlxtn lqj tfcw ouxuwbrt, fxkd $f(x)=x^3-2x^{3/2}-1$ .

$\begin{aligned}f'(1)&=3(1)^2-3\sqrt{1}=0,\\f(1)&=-2.\end{aligned}$

Mlg hrihmhu ya $(1,-2)$ ek $y=-2$ . Up lx slsqytfw xi, kko jlboxtil xuwg, uoh $x$ -hnbb, bd od bok lp $x$ -mjgidpzgs.

Agv iuwsxghiynn jo uoj Qaqdsw-Vfikoms dxudgrv ek fvpb. Qy mwpo xmmtbajxgrlle utw eg rfwtrsjoue tutp $x_0=1$ .

Ikbh Alj: Ni ulrhkke egbspon re t yealb mlhcicym gizqy, zcmkomzgz $f'(x_0)$ pop ndycvgy qkn lybcox cf aon znzbjer. Q fyxdjhtasm oujdg wagyb du fef i gyaradkdyobe vpbdvxwghci; ovk ynjv ovnsljwd, ujzlv vtom wka ziskbwtuzc liypzpb ujfoga kyko qmm btud ekur $f(x_0)\ne0$ .

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